1
Chapter 2 Measurements
2.7Problem Solving
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
A health professional obtains a measured volume from a vial for an injection.
2
To solve a problem• identify the given unit• identify the needed unitProblem:
A person has a height of 2.0 m. What is thatheight in inches?
The given unit is the initial unit of height. given unit = meters (m)
The needed unit is the unit for the answer. needed unit = inches (in.)
Given and Needed Units
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
3
Learning Check
An injured person loses 0.30 pints of blood. Howmany milliliters of blood would that be?
Identify the given and needed units in thisproblem.
Given unit = _______
Needed unit = _______
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
4
Solution
An injured person loses 0.30 pints of blood. Howmany milliliters of blood would that be?
Identify the given and needed units in thisproblem
Given unit = pints Needed unit = milliliters
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
5
STEP 1 State the given and needed quantities.
STEP 2 Write a plan to convert the given unit to the
needed unit.
STEP 3 State the equalities and conversion factors needed to cancel units.
STEP 4 Set up problem to cancel units and calculate answer.
Unit 1 x Unit 2 = Unit 2Unit 1
Given Conversion Needed unit factor unit
Guide to Problem Solving Using Conversion Factors
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
Guide to Problem Solving
6 Basic Chemistry Copyright © 2011 Pearson Education, Inc.
The steps in the Guides to Problem Solving Using Conversion Factors are useful in setting up a problem with conversion factors.
7
Setting up a Problem
How many minutes are 2.5 hours?
STEP 1 Given 2.5 h
Need min
STEP 2 Plan hours minutes
STEP 3 Equalities 1 h = 60 min
STEP 4 Set up problem
2.5 h x 60 min = 150 min
1 h (2 SFs) given conversion needed unit factor unit
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
8
A rattlesnake is 2.44 m long. How long is the snake
in centimeters?
1) 2440 cm
2) 244 cm
3) 24.4 cm
Learning Check
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
9
A rattlesnake is 2.44 m long. How long is thesnake in centimeters?STEP 1 Given 2.44 m
Need centimeters
STEP 2 Plan meters centimeters
STEP 3 Equality 1 m = 100 cm
Factors 1 m and 100 cm
100 cm 1 m
STEP 4 Set up problem
2.44 m x 100 cm = 244 cm (answer 2)
1 m
Solution
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
10
• Often, two or more conversion factors are required to obtain the unit needed for the answer.
Unit 1 Unit 2 Unit 3
• Additional conversion factors are placed in the setup to cancel each preceding unit.Given unit x factor 1 x factor 2 = needed unitUnit 1 x Unit 2 x Unit 3 = Unit 3
Unit 1 Unit 2
Using Two or More Factors
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
11
How many minutes are in 1.4 days?
STEP 1 Given 1.4 days Need minutes
STEP 2 Plan days hours minutes
STEP 3 Equalties: 1 day = 24 h
1 h = 60 minSTEP 4 Set up problem:
1.4 days x 24 h x 60 min = 2.0 x 103 min
1 day 1 h
2 SFs Exact Exact = 2 SFs
Example: Problem Solving
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
12
• Be sure to check your unit cancellation in the setup.• The units in the conversion factors must cancel to
give the correct unit for the answer.
Example:
What is wrong with the following setup?1.4 day x 1 day x 1 h
24 h 60 min
Units = day2/min, which is not the unit needed.
Units do not cancel properly; the setup is wrong.
Check the Unit Cancellation
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
13
Using the GPS
What is 165 lb in kilograms?
STEP 1 Given 165 lb Need kg
STEP 2 Plan lb kg
STEP 3 Equalities/conversion factors
1 kg = 2.20 lb 2.20 lb and 1 kg
1 kg 2.20 lb
STEP 4 Set up problem
165 lb x 1 kg = 74.8 kg (3 SFs) 2.20 lb
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
14
A bucket contains 4.65 L water. How many gallons of water is that?
Learning Check
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
15
STEP 1 Given 4.65 L Need gallons
STEP 2 Plan liters quarts gallons
STEP 3 Equalities
1 L = 1.057 qt 1 gal = 4 qt
STEP 4 Set Up Problem
4.65 L x 1.057 qt x 1 gal = 1.23 gal
1 L 4 qt 3 SFs 4 SFs Exact 3 SFs
Solution
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
16
If a ski pole is 3.0 feet in length, how long is the ski pole in mm?
Learning Check
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
17
STEP 1 Given 3.0 ft Need mm
STEP 2 Plan ft in. cm mm
STEP 3 Equalities 1 ft = 12 in., 2.54 cm = 1 in.,
1 cm = 10 mm
STEP 4 Set up problem
3.0 ft x 12 in. x 2.54 cm x 10 mm = 910 mm (2SFs)
1 ft 1 in. 1 cm
Check initial unit: ft
Check factor setup: units cancel properly
Check needed unit: mm
Solution
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
18
If your pace on a treadmill is 65 meters per
minute, how many minutes will it take for you to
walk a distance of 7500 feet?
Learning Check
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
19
STEP 1 Given 7500 ft; 65 m/min Need min
STEP 2 Plan ft in. cm m min
STEP 3 Equalities 1 ft = 12 in. 1 in. = 2.54 cm
1 m = 100 cm
1 min = 65 m (walking pace)
STEP 4 Set Up Problem7500 ft x 12 in. x 2.54 cm x 1m x 1 min
1 ft 1 in. 100 cm 65 m
= 35 min Answer rounded off (2 SFs)
Solution
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
20
Percent Factor in a Problem
If the thickness of the skin fold at the waist indicates 11% of body fat, how much fat, in kg, does a person have with a body mass of 86 kg?
STEP 1 Given 86 kg; 11% fat Need kg fat
STEP 2 Plan kg body mass kg fat
STEP 3 Equalities/Conversion factors 100 kg of body mass = 11 kg of fat 100 kg body mass and 11 kg fat
11 kg fat 100 kg body massSTEP 4 Set Up Problem
86 kg mass x 11 kg fat = 9.5 kg of fat 100 kg mass
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
21
How many lb of sugar are in 120 g of candy if the
candy is 25% (by mass) sugar?
Learning Check
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
22
STEP 1 Given 120 g of candy
Need lb of sugar
STEP 2 Plan g lb of candy lb of sugar
STEP 3 Equalities 1 lb = 453.6 g
100 g of candy = 25 g of sugar
STEP 4 Set Up Problem % factor
120 g candy x 1 lb candy x 25 lb sugar
453.6 g candy 100 lb candy
= 0.066 lb of sugar
Solution
Basic Chemistry Copyright © 2011 Pearson Education, Inc.